If x=aleft[cos theta+log left{tan left(frac{t | vedclass

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(B) Given $x=a[\cos 	heta+\log (	an (	heta/2))]$ and $y=a \sin 	heta$.First,differentiate $x$ with respect to $	heta$:$\frac{dx}{d	heta} = a \le

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$	an 	heta$

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(B) Given $x=a[\cos 	heta+\log (	an (	heta/2))]$ and $y=a \sin 	heta$.First,differentiate $x$ with respect to $	heta$:$\frac{dx}{d	heta} = a \left[ -\sin 	heta + \frac{1}{	an(	heta/2)} \cdot \sec^2(	heta/2) \cdot \frac{1}{2} ight]$Using the identity $	an(	heta/2) = \frac{\sin(	heta/2)}{\cos(	heta/2)}$ and $\sec^2(	heta/2) = \frac{1}{\cos^2(	heta/2)}$,we get:$\frac{dx}{d	heta} = a \left[ -\sin 	heta + \frac{\cos(	heta/2)}{\sin(	heta/2)} \cdot \frac{1}{\cos^2(	heta/2)} \cdot \frac{1}{2} ight]$$\frac{dx}{d	heta} = a \left[ -\sin 	heta + \frac{1}{2\sin(	heta/2)\cos(	heta/2)} ight]$Since $2\sin(	heta/2)\cos(	heta/2) = \sin 	heta$,we have:$\frac{dx}{d	heta} = a \left[ -\sin 	heta + \frac{1}{\sin 	heta} ight] = a \left[ \frac{1-\sin^2 	heta}{\sin 	heta} ight] = \frac{a \cos^2 	heta}{\sin 	heta} \quad \dots(i)$Now,differentiate $y$ with respect to $	heta$:$\frac{dy}{d	heta} = a \cos 	heta \quad \dots(ii)$To find $\frac{dy}{dx}$,divide equation $(ii)$ by equation $(i)$:$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \cos 	heta}{a \cos^2 	heta / \sin 	heta} = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.

If $x=a\left[\cos \theta+\log \left\{\tan \left(\frac{\theta}{2}\right)\right\}\right]$ and $y=a \sin \theta$,then $\frac{dy}{dx}$ is equal to

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